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solvefor x,y=In[tan(5x^2+3x+2)]

Solución

resolver para

x, y = I n [tan (5 x^{2} + 3 x + 2)]

Solución

x = \frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}, x = \frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}

Pasos de solución

y = I n [tan (5 x^{2} + 3 x + 2)]

Intercambiar lados

I n [tan (5 x^{2} + 3 x + 2)] = y

Dividir ambos lados entre

I n; n \neq = 0

I n [tan (5 x^{2} + 3 x + 2)] = y

Dividir ambos lados entre

I n; n \neq = 0

\frac{I n [ tan ( 5 x ^{2} + 3 x + 2 ) ]}{I n} = \frac{y}{I n}; n \neq = 0

Simplificar

tan (5 x^{2} + 3 x + 2) = \frac{y}{I n}; n \neq = 0

tan (5 x^{2} + 3 x + 2) = \frac{y}{I n}; n \neq = 0

Aplicar propiedades trigonométricas inversas

tan (5 x^{2} + 3 x + 2) = \frac{y}{I n}

Soluciones generales para

tan (5 x^{2} + 3 x + 2) = \frac{y}{I n}

tan (x) = a \Rightarrow x = arctan (a) + πk

5 x^{2} + 3 x + 2 = arctan (\frac{y}{I n}) + πk

5 x^{2} + 3 x + 2 = arctan (\frac{y}{I n}) + πk

Resolver

5 x^{2} + 3 x + 2 = arctan (\frac{y}{I n}) + πk : x = \frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}, x = \frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}

5 x^{2} + 3 x + 2 = arctan (\frac{y}{I n}) + πk

Mover

πk

al lado izquierdo

5 x^{2} + 3 x + 2 = arctan (\frac{y}{I n}) + πk

Restar

πk

de ambos lados

5 x^{2} + 3 x + 2 - πk = arctan (\frac{y}{I n}) + πk - πk

Simplificar

5 x^{2} + 3 x + 2 - πk = arctan (\frac{y}{I n})

5 x^{2} + 3 x + 2 - πk = arctan (\frac{y}{I n})

Mover

arctan (\frac{y}{I n})

al lado izquierdo

5 x^{2} + 3 x + 2 - πk = arctan (\frac{y}{I n})

Restar

arctan (\frac{y}{I n})

de ambos lados

5 x^{2} + 3 x + 2 - πk - arctan (\frac{y}{I n}) = arctan (\frac{y}{I n}) - arctan (\frac{y}{I n})

Simplificar

5 x^{2} + 3 x + 2 - πk - arctan (\frac{y}{I n}) = 0

5 x^{2} + 3 x + 2 - πk - arctan (\frac{y}{I n}) = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

5 x^{2} + 3 x + 2 - πk - arctan (\frac{y}{I n}) = 0

Formula general para ecuaciones de segundo grado:

Para

a = 5, b = 3, c = 2 - πk - arctan (\frac{y}{I n})

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( 2 - πk - arctan ( \frac{y}{I n} ) )}{2 \cdot 5}

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( 2 - πk - arctan ( \frac{y}{I n} ) )}{2 \cdot 5}

Simplificar

3^{2} - 4 \cdot 5 (2 - πk - arctan (\frac{y}{I n})) : 20 πk + 20 arctan (\frac{y}{n I}) - 31

3^{2} - 4 \cdot 5 (2 - πk - arctan (\frac{y}{I n}))

Multiplicar los numeros:

4 \cdot 5 = 20

= 3^{2} - 20 (2 - πk - arctan (\frac{y}{n I}))

3^{2} = 9

= 9 - 20 (2 - πk - arctan (\frac{y}{n I}))

Expandir

9 - 20 (2 - πk - arctan (\frac{y}{I n})) : 20 πk + 20 arctan (\frac{y}{n I}) - 31

9 - 20 (2 - πk - arctan (\frac{y}{I n}))

= 9 - 20 (2 - πk - arctan (\frac{y}{n I}))

Expandir

- 20 (2 - πk - arctan (\frac{y}{I n})) : - 40 + 20 πk + 20 arctan (\frac{y}{n I})

- 20 (2 - πk - arctan (\frac{y}{I n}))

Aplicar la siguiente regla de productos notables

= (- 20) \cdot 2 + (- 20) (- πk) + (- 20) (- arctan (\frac{y}{I n}))

Aplicar las reglas de los signos

+ (- a) = - a, (- a) (- b) = ab

= - 20 \cdot 2 + 20 πk + 20 arctan (\frac{y}{n I})

Multiplicar los numeros:

20 \cdot 2 = 40

= - 40 + 20 πk + 20 arctan (\frac{y}{n I})

= 9 - 40 + 20 πk + 20 arctan (\frac{y}{n I})

Restar:

9 - 40 = - 31

= 20 πk + 20 arctan (\frac{y}{n I}) - 31

= 20 πk + 20 arctan (\frac{y}{n I}) - 31

x_{1, 2} = \frac{- 3 \pm 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{2 \cdot 5}

Separar las soluciones

x_{1} = \frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{2 \cdot 5}, x_{2} = \frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{2 \cdot 5}

x = \frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{2 \cdot 5} : \frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}

\frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{2 \cdot 5}

Multiplicar los numeros:

2 \cdot 5 = 10

= \frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}

x = \frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{2 \cdot 5} : \frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}

\frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{2 \cdot 5}

Multiplicar los numeros:

2 \cdot 5 = 10

= \frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}

Las soluciones a la ecuación de segundo grado son:

x = \frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}, x = \frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}

x = \frac{- 3 + 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}, x = \frac{- 3 - 20 πk + 20 arctan ( \frac{y}{n I} ) - 31}{10}

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